If G is an abelian group and n$\in \mathbb{Z}$, show that $H_n=\{g\in G:g=x^n\ $ for some $x\in G \}$ is a subgroup of G . Give an example where this fails if G is not abelian.
I know the fist part to show $H_n$ is a subgroup. But I failed to find the example. I have tried Klein Four group , Dihedral group. Still can't find such a non abelian group. Why can non abelian group be so rare.
Best Answer
Your second comment was right. $H_2$ does in fact form a subgroup in $D_6$. This is a fluke, and it need not happen.
In $D_6$ it is $H_3$ that goes wrong. If you work out all cubes (not hard once you've already worked out the squares, since $a^3=(a^2)a$), you get $$ H_3 =\{ e, s, rs, r^2s\}. $$ To do this you need the defining relation $srs=r^{-1}$ (and its cousin $sr=r^{-1}s$ is useful for computing the cube of $r^2s$). But this is not a subgroup as it is not closed under the inherited operation: $$ s \cdot rs = srs = r^{-1}=r^2 \notin H_3. $$ Or, if you already have Lagrange at your disposal, it cannot be a subgroup as it has order $4$ inside a group of order $6$.